JEE Maths Quiz on Differential Equations
JEE Maths Quiz on Differential Equations : In this article you will get to Online test for JEE Main, JEE Advanced, UPSEE, WBJEE and other engineering entrance examinations that will help the students in their preparation. These tests are free of cost and will useful in performance and inculcating knowledge. In this post we are providing you quiz on Differential Equations.
Quiz on Differential Equations
Q1. If x dy = y (dx + y dy), y > 0 and y (1) = 1, then y (−3) is equal to _________.
a) 0
b) 3
c) 5
d) 3
b) 3
Solution:
x dy = y (dx + y dy)
[x dy − ydx] / [y^{2}] = dy
−d (x / y) = dy
Integrating both sides, we get x / y + y = c [Because y (1) = 1 ⇒ c = 2; Hence xy + y = 2 ]
For x = −3,
y^{2 }− 2y − 3 = 0
⇒ y = −1 or 3
⇒y = 3 (Because y > 0)
Q2. (x^{2 }+ y^{2}) dy = xy dx. If y (x_{0}) = e, y (1) = 1, then the value of x_{0} = __________.
a) ± √2e
b) ± √3e
c) ± √4e
d) ± √5e
b) ± √3e
Solution:
(x^{2 }+ y^{2}) dy = xy dx
x (x dy − y dx) = −y^{2} dy
[x * (y dx − x dy)] / y^{2} = dy
[x / y] d (x / y) = dy / y
Integrating, x^{2 }/ 2y^{2} = loge y + c
Given y (1) = 1
c = 1 / 2
x^{2 }/ 2y^{2 }= loge y + 1 / 2
Now y (x_{0}) = e
x_{0}^{2 }/ 2e^{2 }− loge e − 1 / 2 = 0
x_{0}^{2 }= 3e^{2}
x_{0 }= ± √3e
Q3. Solution of differential equation 2xy dy / dx = x^{2} + 3y^{2} is __________.
a) x^{2} + y^{2 }= px^{2}
b) x^{2} – y^{2 }= px^{3}
c) x^{3} + y^{3 }= px^{2}
d) x^{2} + y^{2 }= px^{3}
d) x^{2} + y^{2 }= px^{3}
It is homogeneous equation dy / dx = [x^{2} + 3y^{2}] / 2xy
Put y = vx and dy / dx = v + x * [dv / dx]
So, we get x * [dv / dx] = [1 + v^{2}] / 2v
2v dv / 1 + v^{2} = dx / x
On integrating, we get
x^{2} + y^{2 }= px^{3}. (where p is a constant)
Q4. An integrating factor for the differential equation (1 + y^{2}) dx − (tan^{−1 }y − x) dy = 0
a) e^{tan−1y}
b) e^{tan+1y}
c) e^{tan−2y}
d) e^{tan+2y}
a) e^{tan−1y}
Solution:
(1 + y^{2}) dx − (tan^{−1 }y − x) dy = 0
dy / dx = (1 + y^{2}) / (tan^{−1 }y − x)
dx / dy = [tan^{−1 }y / 1 + y^{2}] − [x / 1 + y^{2}] [dx / dy] + [x / 1 + y^{2}] = [tan^{−1 }y] / [1 + y^{2}]
This is equation of the form dx / dy + Px = Q
So, I.F. = e^{∫Pdy}
= e^{∫1 / 1 + y2.dy}
= e^{tan−1y}
Q5. What is the general solution of the differential equation (2x − y + 1) dx + (2y − x + 1) dy = 0?
Solution:
(2x −y + 1) dx + (2y − x + 1) dy = 0
dy / dx = 2x − y + 1x − 2y −1, put x = X + h, y = Y + k
dY / dX = [2X − Y + 2h − k + 1] / [X − 2Y + h −2k − 1]
2h − k + 1 = 0
h − 2k − 1 = 0
On solving h = −1, k = −1;
dY/ dX = [2X − Y] / [X − 2Y]
Put Y = vX;
dY / dX = v + [X dv / dX]
v + [X dv / dX] = [2X − vX] / [X − 2vX] = [2 − v] / [1 − 2v]
X dv / dX = [2 − 2v + 2v^{2}] / [1 − 2v] = 2 (v^{2} −v + 1) / [1 − 2v]
dX / X = (1 − 2v) / 2 (v^{2 }− v + 1) dv
Put v^{2 }− v + 1 = t
(2v − 1) dv = dt
dX / X = −dt / 2t
log X = logt^{− ½ }+ log c
X = t^{−1/2 }c
X = (v^{2} − v + 1)^{−1/2 }* c
X^{2 }(v^{2} − v + 1) = constant
(x + 1)^{2 }(([y + 1)^{2 }/ (x + 1)^{2}] − [(y+1) / (x + 1)] + 1) = constant
(y + 1)^{2 }− (y + 1) (x + 1) + (x + 1)^{2 }= c
y^{2} + x^{2} − xy + x + y = c
Q6. If y′= [x − y] / [x + y], then its solution is ____________.
Solution:
Given dy / dx = x −y / x + y.
Put y = vx
dy / dx = v + x * [dv / dx]
v + x * [dv / dx] = [x − vx] / [x + vx]
v + x [dv / dx] = [1 − v] / [1 + v] [1 + v] / [2 − (1 + v)^{2}] dv = dx / x
Integrating both sides, ∫[1 + v] / [2 − (1 + v)^{2}] dv = ∫dx / x
Put (1 + v)^{2 }= t
⇒ 2 (1 + v) dv = dt
[1 / 2] ∫dt / [2 − t] =∫dx / x
[− 1 / 2] log (2 − t) = log xc
[−1 / 2] log [2 − (1 + v)^{2}] = log xc
[−1 / 2] log [−v^{2 }− 2v + 1] = log xc
log 1 / √[1 − 2v −v^{2}] = log xc
x^{2}c^{2 }(1 − 2v −v^{2}) = 1
y^{2} + 2xy − x^{2} = c_{1}.
Q7. A function y = f (x) has the secondorder derivatives f′′(x) = 6 (x − 1). If its graph passes through the point (2, 1) and at that point the tangent to the graph is y = 3x − 5, then the function is __________.
a) (x + 1)^{3}
b) (x − 1)^{2}
c) (x − 1)^{3}
d) (x + 1)^{2}
c) (x − 1)^{3}
Solution:
Given f′′(x) = 6 (x − 1)
f′(x) = 3 (x − 1)^{2 }+ c_{1} ..(i)
But at point (2, 1) the line y = 3x − 5 is tangent to the graph y = f(x).
Hence, dy / dx∣_{x = 2 }= 3 or
f′ (2) = 3.
Then from (i) f′ (2) = 3 (2 − 1)^{2 }+ c_{1}
3 = 3 + c_{1}
c_{1 }= 0 i.e.,
f′ (x) = 3 (x − 1)^{2}
Given f (2) = 1
f (x) = (x − 1)^{3 }+ c_{2}
f (2) = 1 + c_{2}
1 = 1 + c_{2}
c_{2 }= 0
Hence, f (x) = (x − 1)^{3}.
Q8. The general solution of the differential equation (x + y) dx + x dy = 0 is _______.
Solution:
(x + y) dx + x dy = 0
x dy = − (x + y) dx
dy / dx = −[x + y] / x
It is homogeneous equation, hence, put y = vx and dy / dx = v + x [dv / dx], we get
v + x dv / dx = −[x + vx] / x= −[1 + v] / 1
x dv / dx = −1 −2v
∫dv / [1 + 2v] = −∫dx / x
[1 / 2] log (1 + 2v) = −log x + log c
log (1 + 2 [y / x]) = 2 log [c / x] [x + 2y] / x = (c / x)^{2}
x^{2 }+ 2xy = c.
Q9. Equation of curve through the point (1, 0) which satisfies the differential equation (1 + y^{2}) dx − xy dy = 0, is ________.
a) 3
b) 5
c) 2
d) 1
d) 1
We have [dx / x] = [y dy] / [1 + y^{2}]
Integrating, we get log x = ([1 / 2] * log [1 + y^{2}]) + log c
x = c √(1 + y^{2})−−−−−−−
But it passes through (1, 0), so we get c = 1
Therefore, the solution is x^{2} = y^{2} + 1 or x^{2} − y^{2 }= 1.
Q10. The equation of the curve which passes through the point (1, 1) and whose slope is given by 2y / x, is __________.
a) x^{2}
b) 2x^{2}
c) 1/2x^{2}
d) x^{3}
a) x^{2}
Slope dy / dx = 2y / x
2 ∫dx / x =∫dy / y
2 log x = log y + log c
x^{2 }= yc
Since it passes through (1, 1), therefore, c = 1.
Hence, x^{2 }− y = 0
y = x^{2}.
Subjectwise Tricks Tips & Question with Solution PDFs
S.NO  Subject Name  Topicwise PDFs Download Link 
1.  Chemistry Notes PDF  Click Here to Download Now 
2.  Maths Notes PDF  Click Here to Download Now 
3.  Physics Notes PDF  Click Here to Download Now 
4.  Biology Notes PDF  Click Here to Download Now 
Syllabus and Previous Year Papers 

Chemistry Syllabus for NEET & AIIMS Exams  Click Here 
Chemistry Syllabus for JEE Mains & Advanced  Click Here 
Chapter Wise NEET Chemistry Syllabus  Click Here 
Physics Syllabus for NEET & AIIMS Exams  Click Here 
Physics Syllabus for JEE Mains & Advanced  Click Here 
Chapter Wise NEET Physics Syllabus  Click Here 
Biology Syllabus for NEET & AIIMS Exams  Click Here 
Chapter Wise NEET Biology Syllabus  Click Here 
Maths Syllabus for JEE Mains & Advanced  Click Here 
Download NEET Previous Year Question Papers with Solution  Click Here 


https://www.facebook.com/ExamsRoadOfficial  
Telegram  https://telegram.me/ExamsRoad 
https://twitter.com/ExamsRoad  
https://www.instagram.com/ExamsRoad/  
YouTube  Click Here To Subscribe Now 
Thank You.
By TEAM ExamsRoad.com